# Question #7f415

Apr 14, 2017

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left(x - 1\right)}^{n + 1} / \left(n + 1\right)$

#### Explanation:

Let $g \left(t\right) = \ln \left(1 + t\right) R i g h t a r r o w g \left(0\right) = 0$

By differentiating w.r.t. $t$,

$g ' \left(t\right) = \frac{1}{1 + t} = \frac{1}{1 - \left(- t\right)}$

By the sum of geometric series $\frac{a}{1 - r} = {\sum}_{n = 0}^{\infty} a {r}^{n}$,

$g ' \left(t\right) = {\sum}_{n = 0}^{\infty} 1 \cdot {\left(- t\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{n}$

By integrating w.r.t. $t$,

$g \left(t\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{n + 1} / \left(n + 1\right) + C$,

where $C = 0$ since $g \left(0\right) = 0$

So, we have

$\ln \left(1 + t\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{n + 1} / \left(n + 1\right)$

By setting $t = x - 1$,

$\ln \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left(x - 1\right)}^{n + 1} / \left(n + 1\right)$

I hope that this was clear.